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The magnetic flux through the loop shown in Fig.  increases according to the relation  \phi _B = 6.0t ^2 + 7.0t, where \phi _B is in milliwebers and t is in seconds.  What is the magnitude of the emf induced in the loop when t = 2.0 s?

                                                                  

Option: 1

-(1.6 \times 10^{-2}) \ \frac{wb}{s}


Option: 2

(-3.1 \times 10^{-3}) \ \frac{Wb}{s}


Option: 3

-(1.6 \times 10^{-3}) \ \frac{wb}{s}


Option: 4

-(3.1 \times 10^{-2}) \ \frac{wb}{s}


Answers (1)

best_answer

 

 

Maxwell's equations -

Maxwell's equations

The four Maxwell's equations and Lorentz force law together constitute the foundations of classical electromagnetism. The Maxwell's equations are:

                                \begin{array}{ll}{\text { 1. } \oint \mathbf{E} \cdot \mathrm{d} \mathbf{A}=\mathrm{Q} / \varepsilon_{0}} & {\text { (Gauss's Law for electricity) }} \\ \\ {\text { 2. } \oint \mathbf{B} \cdot \mathrm{d} \mathbf{A}=0} & {\text { (Gauss's Law for magnetism) }} \\ \\ {\text { 3. } \oint \mathbf{E} \cdot \mathrm{d} \mathbf{l}=\frac{-\mathrm{d} \phi_{\mathrm{B}}}{\mathrm{d} t}} & {\text { (Faraday's Law) }} \\ \\ {\text { 4. } \oint \mathbf{B} \cdot \mathrm{d} \mathbf{l}=\mu_{0} i_{\mathrm{c}}+\mu_{0} \varepsilon_{0} \frac{\mathrm{d} \phi_{E}}{\mathrm{d} t}} & {\text { (Ampere-Maxwell Law) }}\end{array}

 

- As we know from faraday's law : 

\varepsilon _{ind}=\frac{-d \phi_{B}}{dt}

\frac{d \phi_{B}}{dt}=\frac{d}{dt}(6t^2+7t)=(12t+7) \ \frac{mWb}{s} 

                         =(12t+7) \times 10^{-3} \ \frac{Wb}{s}

At t=2 sec, 

\varepsilon _{ind}=\frac{-d \phi_{B}}{dt}=-31 \times 10^{-3} \ \frac{Wb}{s}

Therefore \varepsilon _{ind}=-3.1 \times 10^{-2} \ \frac{Wb}{s}

Correct option is (4). 

Posted by

Ritika Kankaria

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